Everything is correct except this: P(AnB) = P(A)*(B) = (1/10)*(2/5) = 2/50
P(AnB) is given by the product of individual probabilities when the events are independent.
How many Chemistry books are hardcover? There are total 8 hardcovers and 7 belong to Math and Physics. This means there is only 1 Chemistry hardcover book out of total 20 books.
So P(AnB) = 1/20

Now P(AuB) = 1/10 + 2/5 - 1/20 = 9/20
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First phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf. To do so, you have the equations:

m + p + c = 20 (since 4/5 of the 25 spots are full of books)

m = 2p

p = 4 + c

From that, you can use Substitution to get everything down to one variable.

c = p - 4

m = 2p

p = p

Then (p - 4) + 2p + p = 20, so 4p = 24 and p = 6. That means that there are 12 math, 6 physics, and 2 chemistry books on the shelf.

With those numbers, you also know that there are 8 total hardcovers, 1 of which is chemistry. So if your goal is to get either a hardcover or a chemistry, there are 9 ways to "win" - either one of the 7 hardcovers that aren't chemistry or the two chemistry books. So out of the 20 total, 9 provide the desired outcome, making the answer E.

Note - a common trap answer here is A, as people forget that "2 chem + 8 hard" double counts the one book that is "both". Venn Diagram logic can be helpful to avoid that trap, or in cases like this with relatively small numbers it may be even more convenient to just write out the possibilities.

the shelf has the capacity of 25 books, 20% empty (5 books), the shelf has 20 books

Phys = 1x
Math = 2x
chem = x - 4

1x + 2x + (x - 4) = 20
4x - 4 = 20
4x = 24
x = 6
x - 4 = 2

So the shelf only has 12 math, 6 phys, 2 chem books

If there are 12 soft covers and 8 hard covers, and 4 hard covers are math/phys, then 4 hard cover must be chemistry, BUT there are only 2 chemistry books.

Please help me with this one.

Maths, Physics and chemistry books are stored on a library shelf that can accomodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books. Among all the books, 12 books are soft cover and the remaining are hard-cover. If there are a total of 7 hard-cover books among the maths and physics books. What is the prob, that a book selected at random is either a hard cover book or a chemistry book?

Chemistry books: 2
Physics books: 6
Math books: 12

Sum = 20

P(Chemistry) = 2/20 = 1/10
P(Hard cover) = 8/20 = 4/10 = 2/5

P(AuB) = P(A)+P(B)-P(AnB)

P(AnB) = P(A)*(B) = (1/10)*(2/5) = 2/50

P(AuB) = 5/50 + 20/50 - 2/50 = 23/50

P(AuB) - P(AnB) = 21/50 (since either means that we don't want both)

What is wrong here? (OA: 9/20)

@GDR29,

Thanks for the explanation but I have one doubt.

We know that there are two chemistry books and there is no information like atleast each of Maths, Physics and Checmistry has to be either soft or hard covered. In such case, there will be two chemistry books which may be hard covered.

Please advise.

Hi guys,
After knowing there are 20 books on the shelve, I've just used the "Among all the books, 12 books are soft cover and the remaining are hard-cover" info. Using this info you get that 8/20 books are hard cover... and at least one book has to be chemestry, so the chances of getting either a hard cover book or a chemestry one have to be higher than 8/20 --- only alternative is E) 9/20
What do you think?
Thanks!

There is a typo in the question given. You are given that there are 12 softcover books (which means there are 8 hard cover books). You also given that 7 of the hardcover books are either Math or Physics. So there is only 1 Chemistry hard cover book.
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That's correct. Note that we do analyze the rest of the data to say that the required probability has to be greater than 8/20 since it is possible that all Chem books are hardcover and hence already a part of the 8 books. But, to say that the required probability is "greater than or equal to" 8/20, all we need is "Among all the books, 12 books are soft cover and the remaining are hard-cover".

There is only one option greater than or equal to 8/20.
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Hello
I am not good at probability, but somehow got this answer right . Could you please explain if the approach used by me is correct or it was just a stroke of luck that i got it right..

Number of Math+Physics+Chemistry books= 25 - ((20/100)*25) = 20
Hence, since M=2P, P=4+C,
(4+C)+C+2(4+C)=20
Therefore, M=12, C=2, P=6

Probability of getting a chemistry book = 2/20 = 1/10
Probability of getting a Hardcover book = P(getting a math or phy book) * P( getting a hardcover book from math or phy books) = (18/20 )*(7/18) = 7/20

Therefore, P(Getting chemistry or hardcover book) = (1/10)+(7/20) = 9/20

I am really doubtful about the red colored approach. Please guide.

Your approach is correct. Though the red part should be {the probability of getting a hardcover math or physics book}.
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Hi Karishma. request you to please help me by explaining why the solution is also accounting for the book which is both chemistry and hard cover when the question stem clearly asks we need to select only books which are "Either" chem or hard cover.
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In Math, 'either or' is interpreted a little differently - it includes the elements which lie in both sets too. "Either Chem or hardcover" means "Chemistry books or hardcover books or both".
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This is a brilliant way of thinking. I really need to develop such ability to get +750. Thanks for sharing easycheesy
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